Sums and Products of Roots: Practice

Here $a = 6$ and $b = -5$, and the roots add to $-\frac{b}{a}$.

$r + s = -\frac{-5}{6} = \frac{5}{6}$

So the sum is $\tfrac{5}{6}$. The value $-\tfrac{2}{3}$ is the product $\frac{c}{a}$, not the sum.

The roots add to $-\frac{b}{a} = \frac{10}{3}$. Subtract the known root from that total.

$s = \frac{10}{3} - 3 = \frac{1}{3}$

So the other root is $\tfrac{1}{3}$. The product checks it, since $\frac{c}{a} = 1$ and $1 \div 3 = \tfrac{1}{3}$.

The roots add to $-b = 7$, so the other root is $7 - 2 = 5$. The constant $c$ is the product.

$c = rs = 2 \cdot 5 = 10$

So $c = 10$. The value $5$ is the other root, not $c$.

The product of the roots is $\frac{c}{a} = \frac{2}{a}$. Set it equal to $1$.

$\frac{2}{a} = 1 \ \Rightarrow \ a = 2$

So $a = 2$. The product formula, not the sum, is what fixes $a$ here.

The sum of the roots is $-\frac{b}{a} = -\frac{k}{2}$. Set it equal to $-4$.

$-\frac{k}{2} = -4 \ \Rightarrow \ k = 8$

So $k = 8$. Multiplying both sides by $-2$ undoes the division and the sign.

The roots must sum to $-b = 3$ and multiply to $c = -40$. Test each factorization's roots.

$(x - 8)(x + 5): \quad 8 + (-5) = 3, \quad 8 \cdot (-5) = -40$

Only $(x - 8)(x + 5)$ passes both tests. For example $(x - 5)(x + 8)$ has roots $5$ and $-8$, summing to $-3$.

Move everything to one side to reach standard form: $2x^2 - 5x - 12 = 0$. Now $a = 2$ and $b = -5$.

$r + s = -\frac{b}{a} = \frac{5}{2}$

So the sum is $\tfrac{5}{2}$. Rearranging before reading the coefficients is essential.

The roots sum to $2 + (-5) = -3$ and multiply to $(2)(-5) = -10$, so the monic form is $x^2 + 3x - 10$. Multiply through by $3$.

$3\left(x^2 + 3x - 10\right) = 3x^2 + 9x - 30 = 0$

This is also $3(x - 2)(x + 5) = 0$, which has the required roots.

The roots add to $-b = 6$, so the other root is $6 - 1 = 5$. The constant $c$ is the product.

$c = rs = 1 \cdot 5 = 5$

So $c = 5$. Substituting $x = 1$ into $x^2 - 6x + c = 0$ gives the same answer.

Here $a = 4$, $b = -4$, and $c = -3$. Use $r + s = -\frac{b}{a}$ and $rs = \frac{c}{a}$.

$r + s = -\frac{-4}{4} = 1, \qquad rs = \frac{-3}{4} = -\frac{3}{4}$

So the sum is $1$ and the product is $-\tfrac{3}{4}$.

A double root at $3$ means both roots are $3$, so the sum is $3 + 3 = 6$ and the product is $3 \cdot 3 = 9$. Use $x^2 - (r + s)x + rs$.

$x^2 - 6x + 9 = 0$

This is $(x - 3)^2 = 0$, exactly the quadratic with a repeated root at $3$.

The roots sum to $\tfrac{2}{3} + \left(-\tfrac{1}{2}\right) = \tfrac{1}{6}$ and multiply to $\tfrac{2}{3} \cdot \left(-\tfrac{1}{2}\right) = -\tfrac{1}{3}$, so the monic form is $x^2 - \tfrac{1}{6}x - \tfrac{1}{3}$. Multiply through by $6$.

$6\left(x^2 - \tfrac{1}{6}x - \tfrac{1}{3}\right) = 6x^2 - x - 2 = 0$

This factors as $(3x - 2)(2x + 1)$, confirming the roots.